= Solution
For causal functions the <convolution> is $(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$. In a half-plane of absolute convergence, changing variables $u=t-\tau$ in its <Laplace transform> and using <Fubini's theorem> gives
$$
\mathcal L(f*g)(s)=\int_0^\infty\int_0^\infty e^{-s(u+\tau)}f(\tau)g(u)du\,d\tau=F(s)G(s).
$$
This proves the <convolution theorem for Laplace transforms>. Transforming the differential equation gives $(s^2+1)Y(s)=sy(0)+y'(0)+F(s)$. The direct transform pairs $\mathcal L(\cos t)=s/(s^2+1)$ and $\mathcal L(\sin t)=1/(s^2+1)$ then give
$$
\boxed{y(t)=y(0)\cos t+y'(0)\sin t+\int_0^t f(\tau)\sin(t-\tau)d\tau.}
$$
For a continuous forcing, call the last term $w(t)$. The <Leibniz integral rule> gives $w'=\int_0^t f(\tau)\cos(t-\tau)d\tau$ and $w''=f(t)-w(t)$. Also $w(0)=w'(0)=0$. Thus both the original equation and the initial values hold.
For an impulse at positive time $T$, the <Dirac delta> evaluates the convolution, yielding
$$
\boxed{y(t)=y(0)\cos t+y'(0)\sin t+H(t-T)\sin(t-T).}
$$
Here $H$ is the <Heaviside step function>. The displacement remains continuous at $T$, while its derivative jumps upward by one, exactly as required by integrating the differential equation across the unit impulse.
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